Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 27



Work Step by Step

Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Plug the derivatives in equation $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ This implies that $\dfrac{2x+y}{2y+x} \times (-1)=\dfrac{-(2x+y)}{2y+x}$ For the point $(1,2)$, we have $\dfrac{-(2x+y)}{2y+x} =\dfrac{-(2+2)}{2(2)+1} =\dfrac{-4}{5}$
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